Question

Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2=12y$ to the ends of its latus-rectum.

Answer 1

The given equation of the parabola is $x^2=12y$

This is of the form $x^2=4ay$ on comparing,we get

4a=12

$\Rightarrow a=\frac{12}{4}=3\therefore \text{Co-ordinates of the focus S is (0,3) P and Q lies on the parabola}\therefore x^2=12\times 3\Rightarrow x^2=36\Rightarrow x=\pm 6\therefore P(-6,3)andQ(6,3)Now,PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(6+6{)}^2+(3-3{)}^2}=\sqrt{(12{)}^2}=12and,OS=3\therefore Areaof△POQ=\frac{1}{2}\times PQ\times OS=\frac{1}{2}\times 12\times 3=6\times 3=18sq.units.$